The curve y=(x^3)/3 - x^2 -3x 4 has a local maximum point at P and a local minimum point at Q. Determine the equation of the straight line passing through P and Q in the form ax by c=0 where a,b,c are all real numbers

First determine who are those points P and Q: as they are local max and min, the derivative is 0 in P and Q.

y=f(x)=\frac{x^3}{3}-x^2-3x+4
f'(x)= x^2-2x-3=(x+1)(x-3)

thus x=-1, and x=3 to which correspond y=17/3 and y=-5, thus one point is (-1,17/3)
and the other (3,-5). Thus the line passing in P and Q is

\frac{y+5}{y-17/3}=\frac{x-3}{x+1}

which leads to 8x+3y-9=0.